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Commit85dea78

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Update 替换空格.py
replace1方法和replace2方法都是O(n*m)的时间复杂度,这里m是空格的数量,因为list的insert是一个O(n)的复杂度。而list的append是一个O(1)的时间复杂度,除了扩容时的时间损耗,append方法只需要遍历一次数组即可得结果。
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‎Target Offer/替换空格.py

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# -*- coding:utf-8 -*-
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classSolution:
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# s 源字符串
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# 使用append一次遍历即可替换
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# 由于list的append是O(1)的时间复杂度,除了扩容所导致的时间损耗,该算法复杂度为O(n)
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defreplaceSpaceByAppend(self,s):
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string=list(string)
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stringReplace= []
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foriteminstring:
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ifitem==' ':
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stringReplace.append('%')
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stringReplace.append('2')
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stringReplace.append('0')
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else:
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stringReplace.append(item)
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return"".join(stringReplace)
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# 创建新的字符串进行替换
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defreplaceSpace1(self,s):

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